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The burning of gasoline in a car releases about 3.0×104kcal/gal.nIf a car averages 42 km/gal when...

The burning of gasoline in a car releases about 3.0×104kcal/gal.nIf a car averages 42 km/gal when driving 80 km/h , which requires 23 hp , what is the efficiency of the engine under those conditions?

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Answer #1

1 horsepower = 746 watts

23 hp = 17158 watts

1 kcal = 4186 joules (watts / sec)

e = W / Qh = 1 - (Ql / Qh)

1 gal / 42 km at 80 km / h means that that 1.90 gallons will be used in one hour.

Since no temperatues are given 1 - Ql / Qh is not needed...

Converting 30000 kcal / gal to watts I get 34883 joules/sec. Since 1.90 gal are used,

multiplied the above by 1.90

1.90x34883= 66277.7 .

Dividing 17158 watts (from hp) by 66277.7

17158/66277.7

an efficiency = 25.80%.

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